Physical Layer

Data Transmission

Terminology

Data transmission occurs between a transmitter & receiver via some medium
Data
- Entites that convey meaning
Signals & Signalling
- Electric or electromagnetic representations of data, physically propagates along medium
Transmission
- Communication of data by propagation and processing of signals

Data Transmission

Source Data:

Digital data - discrete, e.g. text and integers
Analog data - continuous, e.g. audio

Transmitted Signal:

Digital signal - a sequence of voltage pulses transmitted over a medium
Analog signal - a continuous varying electromagnetic wave

4 Types of Data Transmission

Analog Tranmission

Analog data, analog signal: Amplitude Modulation (AM), FM, PM.
Digital data, analog signal: Amplitude Shift Keying (ASK), FSK, PSK.

Digital Transmission

Digital data, digital signal: NRZ-L, NRZI, Bipolar-AMI, …
Analog data, digital signal: Pulse Code Modulation (PCM)

Transmission Media

Transmission media is the physical medium that carries the transmitted signals.

Types of Transmission Media

Guided media (wired)
Unguided media (wireless)
- transport electromagnetic waves without using a physical conductor

Key concerns are data rate and distance, affect the choice of media.

Design Factors

Bandwidth
- Higher bandwidth gives higher data rate
Transmission impairments
- e.g. attenuation (衰減) , Distortion, Noise
Interference (干擾)
Number of receivers in guided media
- More receivers introduces more attenuation

Transmission Impairments

Signal received may differ from signal transmitted causing

Analog- degradation of signal quality
Digital - bit errors

The Most significant impairments are

Attenuation (衰減)
Distortion
Noise

Since there are problems in the electrical and electromagnetic worlds

Resistance (leads to loss)
Capacitance (leads to distortion)
Inductance (leads to interference)

Random electromagnetic radiation is called noise

Can be generated by specific sources such as electric motor
Background radiation is an inescapable feature of the universe

Attenuation

Attenuation means a loss of energy where signal strength falls off with distance
Depends on medium
Mainly resistance loss
Received signal strength must be:
- Strong enough to be detected
  - Solution: increase strength using amplifiers/repeaters
- Sufficiently higher than noise to receive without error
  - Solution: increase strength using amplifiers/repeaters
- Attenuation varies with frequency
  - Solution: equalize attenuation across a band of frequencies

Delay distortion

Propagation velocity varies with frequency
Hence various frequency components arrive at different times
Particularly critical for digital data

Inter-symbol interference: Parts of one bit spill over into others

Noise

Additional signals inserted between transmitter and receiver

Types of Noise:

Thermal Noise (white noise)
- Due to thermal agitation of electrons
- Uniformly distributed, white noise
Intermodulation Noise
- Signals that are the sum and difference of original frequencies sharing a medium
Crosstalk Noise
- A signal from one line is picked up by another
Impulse Noise
- Irregular pulses or spikes, e.g., external electromagnetic interference
- Short duration, high amplitude
- A minor annoyance for analog signals but a major source of error in digital data
- A noise spike could corrupt many bits

Signal to Noise Ratio (SNR)

$\text{SNR} = \frac{\text{Signal Power}}{\text{Noise Power}} = \frac{\text{Signal Voltage}^2}{\text{Noise Voltage}^2}$

$\text{SNR}_{dB} = 10 log_{10}\text{SNR}$

Channel capacity of a transmission media

Channel Capacity is…

A very important consideration in data communications is how fast we can send data, in bits per second, over a channel.

$\text{Channel capacity} = \text{Maximum possible data rate on communications channel}$

Channel capacity depends on:

bandwidth available
Level of the signals we use
Quality of the channel (the level of noise)

Bandwidth

Bandwidth in hertz(Hz), or cycles per second, refers to:

the range of frequencies in a composite signal

the range of frequencies that a channel can pass

The bandwidth is different for each type of transmission medium, e.g. twisted pair (300 kHz), coaxial cable (500 MHz), fiber optics (10 GHz)

In computer networking, we use the term ”bandwidth” in a different context, i.e., bandwidth in bits per second (bps), refers to the speed of bit transmission in a channel or link.

Baud rate (signaling rate)

number of signal elements per second on a channel

Bit rate (transmission rate)

Baud rate is not equal to Bit rate.

For a binary level, baud rate = bit rate
For a multi-level, baud rate < bit rate

Noiseless Channel - Nyquist Bandwidth

used to determine the maximum information rate for noiseless channel.

Nyquist Bitrate Formula is given by

$\mathrm{C}=2 \mathrm{B} \log _{2} \mathrm{L} \text { bps }$

Where

$C$ = bitrate in bps
$B$ = bandwidth in Hz
$L$ = no. of levels per signaling element

Noisy Channel - Shannon Capacity

Shannon Capacity Formula is given by

$C = \mathrm{B} \log_{2}(1+\frac{S}{N}) \text { bps }$

Where

$C$ = information rate in bps
$B$ = bandwidth in Hz
$S/N$ = signal to noise ratio (Signal Power / Noise Power) = SNR

Log Base 2 Calculator

For Calculator, you can type

$log(2, Input)$

or log(input) / log(2).

Line coding techniques

Types of Line coding:

Unipolar - NRZ
Polar - NRZ, RZ, and biphase (Manchester, and differential Manchester)
Bipolar - AMI and pseudoternary
Multilevel - 2B/1Q, 8B/6T, and 4D-PAM5
Multitransition - MLT-3

NRZ

Used for magnetic recording

Not often used for signal transmission

Pros of NRZ

Easy to engineer

make good use of bandwidth

Cons of NRZ

dc component

lack of synchronization capability

Unipolar

Unipolar - NRZ-L

Non-Return to Zero-Level

signal changes between bit

Voltage constant during bit interval

For coming bit, the constant is:

“0”: zero voltage
“1”: postive voltage

Polar

Polar - NRZ-L

Non-Return to Zero-Level

signal changes between bit

Voltage constant during bit interval

For coming bit, the constant is:

“0”: postive voltage
“1”: negative voltage

Polar - NRZI

Non-Return to Zero Inverted

signal changes between bit

Constant voltage pulse for duration of bit

For coming bit, the constant is:

“0”: no transition (no change)
“1”: toggle high or low
- need to check preceding bit of 1 (whether the voltage is +ve or -ve)

Polar - RZ

Return to Zero

signal changes during the bit (middle of the bit)

For coming bit, the constant of first half is:

“0”: negative voltage
“1”: postive voltage

Then in the middle of the bit, in second half of bit is 0 voltage.

No DC component if numbers of “1” and “0” are the same

Biphase - Manchester

Has transition in middle of each bit period

Low to high represents 1
High to low represents 0

Biphase - Differential Manchester

Midbit transition is clocking only

For first bit:

Inversion at start of bit period representing 0
No transition at start of bit period representing 1

For coming bit:

“0” : “No transition”, It goes up and down in the first half
“1” : Inversion in the middle

Used by IEEE 802.5

Bipolar

Bipolar-AMI

Alternate Mark Inversion

signal changes between bit

Constant voltage pulse for duration of bit

For coming bit, the constant is:

“0”: no line signal (0)
“1”: toggle high or low
- need to check preceding bit of 1 (whether the voltage is +ve or -ve)

Pseudoternary

very similar to Bipolar-AMI.

signal changes between bit

Constant voltage pulse for duration of bit

For coming bit, the constant is:

“0”: toggle high or low
- need to check preceding bit of 0 (whether the voltage is +ve or -ve)
“1”: no line signal (0)

Multilevel

2B1Q

Two binary, one quaternary (2B1Q)

It uses data patterns of size 2 and encodes the 2-bit patterns as one signal element belonging to a four-level signal.

8B6T

Eight binary, six ternary (8B6T)

To encode 1 pattern of 8-bit as a pattern of six signal elements, where the signal has three levels (ternary).

28 = 256 and 36 = 729, there are 729-256=473 redundant signal elements that provide synchronization and error detection.

The idea is to encode a pattern of 8 bits as a pattern of 6 signals.
The Three Levels: ( - , 0, +)
To create DC balance last bit pattern is inverted using weight -1 by sender.

To Draw the Graph

Convert the bit into hex value

Use to hex value to find the six signal elements in the 8B/6T table (will be given)

Use weight 1, only the last bit pattern use weight -1

4D-PAM5

Four-dimensional five-level pulse amplitude modulation

The 4D means that data is sent over four wires at the same time.
It uses 5 voltage levels: -2, -1, 0, +1, +2.
Level 0 is used only for forward error detection.

Used in Gigabit LANs

Multitransition

MLT-3

uses three levels (+V, 0 , -V) and three transition rules to move between the levels

If the next bit is 0, there is no transition
If the next bit is 1 and the current level is not 0, the next level is 0.
If the next bit is 1 and the current level is 0, the next level is the opposite of the last non-zero level.

Scrambling

Use scrambling to replace sequences that would produce constant voltage
These filling sequences must
- Produce enough transitions to sync
- Be recognized by receiver & replaced with original
- Be same length as original

Design goals of Scrambling

Have no dc component
Have no long sequences of zero level line signal
Have no reduction in data rate
Give error detection capability

Bipolar with 8-zeros Substitution (B8ZS)

B8ZS substitute eight consecutive zeros with 000VB0VB

V means previous non-zero bit.

B means Inverted previous non-zero bit.

If Previous Level is Positive Voltage:

the eight zeros of the octets are encoded as 000±0-+

If Previous Level is Negative Voltage:

the eight zeros of the octets are encoded as 000-+0±

High-density bipolar-3 zeros (HDB3)

HDB3 substitute four consecutive zeros with 000V or B00V

V means previous non-zero bit.

B means Inverted previous non-zero bit.

If the number of nonzero(1) pulses after the last substitution is Even:

Even number of nonzeros: B00V

Odd number of nonzeros: 000V

Key Concept Review Q&A

Review Questions

What is the position of the transmission media in the OSI or the Internet model?

Name the two major categories of transmission media.

How do guided media differ from unguided media?

What are the three major classes of guided media?

Names three types of transmission impairments.

What does the Nyquist theorem have to do with communications?

What does the Shannon Capacity have to do with communications?

Distinguish between baseband transmission and broadband transmission.

List three techniques of digital-to-digital conversion.

Distinguish between a signal element and a data element.

Distinguish between data rate (bit rate) and signal rate (baud rate).

Define a DC component and its effect on digital transmission.

Define the characteristics of a self-synchronizing signal.

Define scrambling and give its purpose.

Problems

Note:

ps: 10^-12 s

ns: 10^-9 s

$\mu$ s: 10^-6 s

ms: 10^-3 s

KB : 10^3 B

MB : 10^6 B

GB : 10^9 B

TB : 10^12 B

(a)

10 / 1000 = 0.01 s

(b)

8 / 1000 = 0.008 s = 8ms

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$(8 \times 100000) / 1000 = 800$ s

$\text{SNR} = \frac{\text{Signal Power}}{\text{Noise Power}} = \frac{\text{Signal Voltage}^2}{\text{Noise Voltage}^2}$

Signal Voltage = 20(Noise Voltage)

Then square both sides

Signal Power = 400(Noise Power)

SNR = 400

$SNR_{dB} = 10 log_{10} SNR = 26.02$ dB

Shannon Capacity Formula is given by

$C = \mathrm{B} \log_{2}(1+\frac{S}{N}) \text { bps }$

Where

$C$ = information rate in bps

$B$ = bandwidth in Hz

$S/N$ = signal to noise ratio (Signal Power / Noise Power) = SNR

Maximum data rate supported by this line:

$4000 \space log_2(1 + 1000) = 39868.9Kbps$

~Around 40 Kbps

Log Base 2 Calculator

For Calculator, you can type

$log(2, Input)$

or log(input) / log(2).

$SNR_{dB} = 10 log_{10} SNR$

Shannon Equation: $C = \mathrm{B} \log_{2}(1+\frac{S}{N}) \text { bps }$

(a.)

$SNR = 10^{(40/10)}$

Bandwidth = 20KHz

Using Shannon Equation:

$C = 20 \log_{2}(1+10^4) \text { bps } = 265.8 Kbps$

(b.)

$SNR = 10^{4/10}$

Bandwidth = 200KHz

Using Shannon Equation:

$C = 200 \log_{2}(1+10^{4/10}) \text { bps } = 362.5 Kbps$

(c.)

$SNR = 10^{(20/10)}$

Bandwidth = 1MHz = 1000KHz

Using Shannon Equation:

$C = 1000 \log_{2}(1+10^{2}) \text { bps } = 6658 Kbps = 6.66 Mbps$

(a)

SNR = $10^{30/10} = 1000$

Bandwidth = 3.4 KHz

Using Shannon Equation:

$C = 3.4 \log_{2}(1+10^{3}) \text { bps } = 33.9 Kbps$

(b)

Minimum SNR:

$9.6 Kbps = 3.4 log_2(1+SNR)$

$2^{9.6/3.4} = (1 + SNR)$

SNR = 6.079

$SNR_{dB} = 10 log_{10} SNR = 7.838$ dB